Problem: Ron started a construction company. The net value of the company (in thousands of dollars) $t$ months after its creation is modeled by $v(t)=2t^2-12t-14$ If a company's net value is $0$ dollars, the company is breaking even. Ron wants to know how many months it will take for his company to break even. 1) Rewrite the function in a different form (factored or vertex) where the answer appears as a number in the equation. $v(t)=$ 2) How many months after its creation does the company break even?
Solution: Choosing a form Ron's company is breaking even when its net value is zero. So we're looking for what values of $t$ make the output of the function $0$. Which form reveals this feature? Here's a summary of what each form reveals along with examples. Note that these are all equivalent forms of the same function, but not the function modeling the value of the company. Form Example Feature revealed Standard $f(x)=2x^2-12x+{10}$ $y$ -intercept is ${10}$ Factored $f(x)=2(x-C{1})(x-C{5})$ Zeros are $x=C1$ and $x=C5$ Vertex $f(x)=2(x-{3})^2{-8}$ Vertex is $(3,{-8})$ Rewrite in factored form The zeros of the function tell us which values of $t$ make the output of the function $0$, so let's rewrite $v(t)$ in factored form: $\begin{aligned} v(t)&=2t^2-12t-14 \\\\ &=2(t^2-6t-7) \\\\ &=2\left(t+1\right)\left(t-7\right) \end{aligned}$ When does the company break even? The factored form of the function reveals its zeros: $\begin{aligned} &0=2\left(t+1\right)\left(t-7\right) \\\\ &t+1=0\text{ or }t-7=0 \\\\ &\xcancel{t=-1} \text{ or }t=7 \end{aligned}$ A negative value for time doesn't make sense in this context, so Ron's company is breaking even at $t=7$ months. Answers 1) The factored form of the function reveals when the company breaks even: $v(t)=2\left(t+1\right)\left(t-7\right)$ 2) The company breaks even $7$ months after creation.